Question

for a gaseous reaction 2A 2 + 5B2 gives 2A2B5 at 27 degree Celsius of heat change at constant pressure is found to be - 50160 joule calculate the value of internal energy change

Answer 1

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Answer 2

The value of internal energy change is - 37689J

Explanation:

The reaction is:

$2A_2(g)+5B_2(g)\rightarrow 2A_2B_5(g)$

$\Delta H=\Delta U+\Delta n_g\times RT$

where,

$\Delta H$ =  heat change at constant pressure = -50160 J

$\Delta U$= heat change at constant volume or internal energy change = ?

$\Delta n_g$ = change in the moles of the reaction = Moles of product - Moles of reactant = 2 - 7 = -5 mole

R = gas constant = 8.314 J/mole.K

T = temperature = $27^0C=(27+273)K=300K$

Putting in the values we get:

$-50160 J=\Delta U+(-5\times 8.314\times 300)$

$-50160J=\Delta U+(-12471J)$

$\Delta U=-37689J$

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